Proof: if $T$ is self-adjoint, then $\text{det}( T-tI_{V})$ splits over $\mathbb{R}$?

Lemma:

If $T$ is self-adjoint on finite dimensional vector space $V$, then $\text{det} (T-tI_{V})$ splits over $\mathbb{R}$.

I would like to know "why"?

Notice: here we use this Lemma for proving the diagonalization theorem or another form of spectral theorem for self-adjoint operators, so we need a direct proof as @RobertLewis provided.

Theorem:

Suppose $T$ is linear transformation on $V$. $T$ is a self-adjoint operator if and only if there exists an orthonormal basis such that $[T]_{\beta}$ is diagonal.

Proof:

($\Rightarrow$) $T$ is self-adjoint. Then, $\text{det} (T-tI_{V})$ splits over $\mathbb{R}$. Then, by Schur theorem there exists an orthonormal basis $\beta$ such that $[T]_{\beta}$ is uppertriangular. We have: $[T]_\beta = ([T]_{\beta})^{*}$. Hence, $[T]_{\beta}$ is diagonal.

...

It is clear that we use the above Lemma to prove this theorem not conversely!

• Spectral theorem – user223391 Nov 7 '17 at 6:24
• @ZacharySelk: we use this Lemma to prove the spectral theorem about self-adjoint operators NOT conversly. – Amin Nov 8 '17 at 1:41

The eigenvalues of a self-adjoint operator such as $T$ are all real: if

$T v = \lambda v, \tag 1$

then

$\lambda \langle v, v \rangle = \langle v, \lambda v \rangle = \langle v, Tv \rangle, \tag 2$

so

$\bar \lambda \langle v, v \rangle = \overline{\langle v, Tv \rangle} = \langle Tv, v \rangle = \langle v, T^\dagger v \rangle = \langle v, Tv \rangle = \lambda \langle v, v \rangle, \tag 3$

whence, since $\langle v, v \rangle \ne 0$,

$\bar \lambda = \lambda \in \Bbb R. \tag 4$

Since the eigenvalues $\lambda_i \in \Bbb R$ of $T$ are the roots of the $\deg(\dim V)$ polynomial

$p_T(t) = \det(T - t I_V), \tag 5$

we have

$p_T(t) = \det(T - tI_V) = \displaystyle \prod_1^{\dim V}(\lambda_i - t), \tag 6$

i.e, $p_T(t)$ is the product of $\dim V$ linear factors $\lambda_i - t \in \Bbb R[t]$; thus $p_T(t)$ splits in $\Bbb R[t]$, that is, over $\Bbb R$.

The "why" is essentially that $T = T^\dagger$ forces all $\dim V$ of the $\lambda_i \in \Bbb R$, leading to one real factor $\lambda_i - t$ for each of the $\dim V$ eigenvalues.

Note: The assumption that $\dim V < \infty$ is used to affirm the existence of $p_T(t) = \det(t - t I_V)$. End of Note.

• Thanks @RobertLewis. I did not pay attention to the equation (6)! – Amin Nov 7 '17 at 6:59
• My pleasure, sir! – Robert Lewis Nov 7 '17 at 7:04